16.36A
Explanation:
We'll begin by writing a balanced dissociation equation of aqueous Cr2(SO4)3. This is illustrated below:
Cr2(SO4)3 —> 2Cr^3+ 3(SO4)^2-
From the above, we can see that Cr is trivalent.
Next, let us determine the number of faraday needed to deposit metallic Cr. This is illustrated below:
Cr^3+ 3e- —> Cr
From the above equation, 3 faradays are needed to deposit metallic Cr
1 faraday = 96500C
Therefore, 3 faraday = 3 x 96500C = 289500C.
Molar Mass of Cr = 52g/mol
Now let us determine the quantity of electricity needed for 2.68g of Cr metal
This is shown below:
52g of Cr required 289500C.
Therefore, 2.68g of Cr will require = (2.68 x 289500)/52 = 14920.38C
Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:
Q (quantity of electricity) = 14920.38C
t (time) = 15.2mins = 15.2 x 60 = 912secs
I (current) =?
Apply the equation Q = It
Q = It
14920.38 = I x 912
Divide both side by 912
I = 14920.38/912
I = 16.36A
Therefore, a current of 16.36A is needed for the process.